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Loser66
 one year ago
Let A, B \(\in M(n, \mathbb C)\)and assume B is invertible. Show that
1) \((B^{1}AB)^k= B^{1}A^kB\)
2) Use 1 to show \(e^{tB^{1}AB}=B^{1}e^{tA}B\)
Please help
Loser66
 one year ago
Let A, B \(\in M(n, \mathbb C)\)and assume B is invertible. Show that 1) \((B^{1}AB)^k= B^{1}A^kB\) 2) Use 1 to show \(e^{tB^{1}AB}=B^{1}e^{tA}B\) Please help

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Loser66
 one year ago
Best ResponseYou've already chosen the best response.01) is easy. But I don't know how to do 2 @ganeshie8 @dan815

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1is it given that A is diagonal matrix ?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0invertible meaning nonsingular matrix.. there's an inverse.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0nvm, I got it. :) Thanks for being here.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0for 1) \((B^{1}AB)^k = (B^{1}AB)(B^{1}AB)........(B^{1}AB)\) open parenthesis and \(BB^{1}=I\) gives us the answer

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0for 2) by definition \(e^{tA}=\sum_{k=0}^{\infty} \dfrac{t^k}{k!}A^k\)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0we replace A by \(B^{1}AB\) then just simplify it to get the answer.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1\(e^{tB^{1}AB}=\sum_{k=0}^{\infty} \dfrac{t^k}{k!}(B^{1}AB)^k =\sum_{k=0}^{\infty} \dfrac{t^k}{k!}B^{1}A^kB \) ?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0yeah.... this proof looks familiar. Miracrown helped me with one of them.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0@ganeshie8 sum and t, k are constants, right? hence, we can distribute inside, between \(B^{1}\) and A

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Are you saying we can take \(B^{1}\) out ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1\(\sum_{k=0}^{\infty} \dfrac{t^k}{k!}B^{1}A^kB =B^{1}\left(\sum_{k=0}^{\infty} \dfrac{t^k}{k!}A^k\right)B \) ? this looks ok intuitively but idk what you're gonna use as justification

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Look at the definition of \(e^{tA}\) above

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1I am asking the justification for that distribution

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1I think its not so obvious, at least for me it is not.

dan815
 one year ago
Best ResponseYou've already chosen the best response.1if u expand that summation u can show these 2 summations are equation therefore

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0hey, barbecue, @ganeshie8 asked whether we can put the sum between B^1 and A or not. You proved \(A(B_1+B_2) = AB_1 +AB_2\) which is irrelevant to the question. ha!!

dan815
 one year ago
Best ResponseYou've already chosen the best response.1\[\sum_{k=0}^{\infty} \dfrac{t^k}{k!}B^{1}A^kB =B^{1}\left(\sum_{k=0}^{\infty} \dfrac{t^k}{k!}A^k\right)B\]

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Oh, yeah, barbecue!! Thanks for making it clear.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0consider \(C=B^{1}AB\). now suppose \(C^{n1}=B^{1}A^{n1}B\). it follows that $$C^n=C\cdot C^{n1}=B^{1}ABB^{1}A^{n1}B=B^{1}AA^{n1}B=B^{1}A^nB$$. so by induction it follows \(C^n=B^{1}A^n B\) define \(e^X=\sum_{n=0}^\infty \frac1{n!}X^n\) in analogy with the real exponential. $$\begin{align*}e^{B^{1}AB}=\sum_{n=0}^\infty\frac1{n!}(B^{1}AB)&=\sum_{n=0}^\infty\frac1{n!}B^{1}A^nB\\&=B^{1}\sum_{n=0}^\infty\frac1{n!}A^n B\\&=B^{1}\left(\sum_{n=0}^\infty\frac1{n!}A^n\right)B\\&=B^{1}e^AB\end{align*}$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the above follows because matrix multiplication distributes over addition: $$B^{1}(A_1+A_2)B=B^{1}(A_1B+A_2B)=B^{1}A_1B+B^{1}A_2B$$
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